Chủ đề 9: Axit clohiđric

1. PTHH: Zn + 2HCl ➝ ZnCl₂ + H₂↑

2. n_Zn = 23/65 = 0,35(mol)

PTHH: Zn + 2HCl ➝ ZnCl₂ + H₂↑

Số mol:0,35------------>0,35--->0,35 (mol)

⇒ m_ZnCl2 = 0,35. 136 = 47,6(g)

3. V_H2(đktc) = 0,35. 22,4 = 7,84 (lít)

a)

\(n_{HCl} = 0,35.2 = 0,7(mol)\\ n_{Mg} = a\ mol ; n_{Al} = b\ mol\)

\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)

Ta có : 

\(\left\{{}\begin{matrix}24a+27b=7,5\\2a+3b=0,7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

Vậy : 

\(\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8\left(gam\right)\\m_{Al}=0,1.27=2,7\left(gam\right)\end{matrix}\right.\)

b)

\(\left\{{}\begin{matrix}n_{MgCl2}=a=0,2\left(mol\right)\\n_{AlCl3}=b=0,1\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{MgCl2}=0,2.95=19\left(gam\right)\\m_{AlCl3}=0,1.133,5=13,35\left(gam\right)\end{matrix}\right.\)

a, Gọi n_Mg = a; n_Al = b (mol)

⇒ 24a + 27b = 7,5 (1)

n_HCl = 0.7 (mol)

Mg⁰ → Mg⁺² + 2e

a ..................... 2a

Al⁰ → Al⁺³ + 3e

b .................... 3b

2H⁺ + 2e → H₂⁰

0,7 .... 0,7

⇒ 2a + 3b = 0,7 (2)

Từ (1), (2) ⇒ \(\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\left(mol\right)\)

⇒ \(\left\{{}\begin{matrix}n_{Mg}=0,2\\n_{Al}=0,1\end{matrix}\right.\left(mol\right)\)

⇒ m_Mg = 4.8 (g)

và m_Al = 2.7 (g)

b, n_{magie clorua} = n_Mg = 0,2 (mol)

⇒ m_{magie clorua} = 19 (g)

n_{nhôm clorua} = n_Al = 0,1 (mol)

⇒ m_{nhôm clorua} = 13,35 (g)

m_{hỗn hợp muối sau phản ứng} = 13, 35 + 19 = 32, 25 (g)

\(Đặt:nMg=a\left(mol\right),n_{MgCO_3}=b\left(mol\right)\)

\(n_{khí}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(a................a.......a\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

\(b....................b........b\)

\(m_X=24a+84b=13.2\left(g\right)\left(1\right)\)

\(n_{khí}=a+b=0.3\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.1\)

\(m_{Mg}=0.2\cdot24=4.8\left(g\right),m_{MgCO_3}=0.1\cdot84=8.4\left(g\right)\)

\(m_{ddB}=13.2+200-0.2\cdot2-0.1\cdot44=208.4\left(g\right)\)

\(m_{MgCl_2}=0.3\cdot95=28.5\left(g\right)\)

\(C\%MgCl_2=\dfrac{28.5}{208.4}\cdot100\%=13.67\%\)

\(n_{H_2} = \dfrac{4,35-3,95}{2} = 0,2(mol)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)

\(\left\{{}\begin{matrix}Mg:x\left(mol\right)\\Al:y\left(mol\right)\end{matrix}\right.\)→ \(\left\{{}\begin{matrix}24x+27y=4,35\\x+1,5y=0,2\end{matrix}\right.\)→\(\left\{{}\begin{matrix}x=0,125\\y=0,05\end{matrix}\right.\)

Vậy :

\(\%m_{Mg} = \dfrac{0,125.24}{4,35}.100\% = 68,97\%\\ \%m_{Al} = 100\% - 68,97\% = 31,03\%\)

a)

Zn + 2HCl  →  ZnCl₂  +  H₂

Fe + 2HCl  →  FeCl₂  +  H₂

b) Gọi số mol Zn và Fe có trong 17,7 gam hỗn hợp là x và y mol. nH₂ = \(\dfrac{6,72}{22,4}\)=0,3 mol

Theo tỉ lệ phản ứng ta có \(\left\{{}\begin{matrix}x+y=0,3\\65x+56y=17,7\end{matrix}\right.\)=> x = 0,1 và y = 0,2 

=> m_Zn = 0,1.65 = 6,5 gam và m_Fe= 0,2.56 = 11,2 gam

c) nHCl = 2nH₂ = 0,3.2 = 0,6 mol

Áp dụng ĐLBT khối lượng => m muối clorua =  mKl + mHCl - mH₂ 

<=> m muối = 17,7 + 0,6.36,5 - 0,3.2 = 28,05 gam

\(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\\n_{Fe}=c\left(mol\right)\end{matrix}\right.\)⇒ 24a + 27b + 56c = 26,05(1)

\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al +6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\)

\(Mg + Cl_2 \xrightarrow{t^o} MgCl_2\\ 2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3\\ 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ n_{Cl_2} = a + 1,5b + 1,5c = \dfrac{17,36}{22,4} = 0,775(3)\)

Từ (1)(2)(3) suy ra:  a = 0,325 ; b = -0,05 ; c = 0,35

→ Sai đề.

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2................................0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(n_{KMnO_4} = \dfrac{15,8}{158} = 0,1(mol)\\ n_{HCl} = 0,08.2 = 0,16(mol)\)

2KMnO₄ + 16HCl \(\to\) 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

0,02.............0,16...................................0,05..............(mol)

\(n_{NaOH} = 0,2.1,5 = 0,3(mol)\)

2NaOH + Cl₂ \(\to\) NaCl + NaClO + H₂O

0,1..........0,05......0,05......0,05........................(mol)

Vậy :

\(C_{M_{NaCl}} = C_{M_{NaClO}} = \dfrac{0,05}{0,2}= 0,25M\\ C_{M_{NaOH}} = \dfrac{0,3-0,1}{0,2} = 1M\)

Đề bài phải là thể tích CO₂ bạn nhé!

a, PT: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

Ta có: \(n_{CaCO_3}=\dfrac{4}{100}=0,04\left(mol\right)\)

\(m_{HCl}=\dfrac{14,6.25}{100}=3,65\left(g\right)\Rightarrow n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,1}{2}\), ta được HCl dư.

Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,04\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,04.22,4=0,896\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{CaCO_3}=0,08\left(mol\right)\\n_{CaCl_2}=n_{CaCO_3}=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,02\left(mol\right)\Rightarrow m_{HCl\left(dư\right)}=0,02.36,5=0,73\left(g\right)\)

\(m_{CaCl_2}=0,04.111=4,44\left(g\right)\)

Bạn tham khảo nhé!

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{4}{100}=0,04\left(mol\right)\\n_{HCl}=\dfrac{14,6\cdot25\%}{36,5}=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,1}{2}\) \(\Rightarrow\) HCl còn dư, CaCO₃ p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{CO_2}=n_{CaCl_2}=0,04\left(mol\right)\\n_{HCl\left(dư\right)}=0,02\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,04\cdot22,4=0,896\left(l\right)\\m_{CaCl_2}=0,04\cdot111=4,44\left(g\right)\\m_{HCl\left(dư\right)}=0,02\cdot36,5=0,73\left(g\right)\end{matrix}\right.\)

\(n_{Fe}=\dfrac{1.12}{56}=0.02\left(mol\right)\)

\(n_{HCl}=0.1\cdot0.3=0.03\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.015.....0.03.......0.015\)

\(m_{Fe\left(dư\right)}=\left(0.02-0.015\right)\cdot56=0.28\left(g\right)\)

\(m_{FeCl_2}=0.015\cdot127=1.905\left(g\right)\)